Cross Product Calculator

About the Cross Product Calculator

The cross product takes two vectors in space and produces a third vector perpendicular to both. It is the standard way to find a normal vector to a plane, compute the area of a parallelogram or triangle in 3D, and describe rotational quantities in physics such as torque and angular momentum.

This calculator computes $\mathbf{u} \times \mathbf{v}$ for three-dimensional vectors using the familiar determinant layout with $\mathbf{i}$ , $\mathbf{j}$ and $\mathbf{k}$ in the first row. It shows each component separately, including the sign change in the middle term that causes most errors, then gives the magnitude and confirms that the result is orthogonal to both inputs.

What is the cross product?

For vectors $\mathbf{u} = \langle u_1, u_2, u_3 \rangle$ and $\mathbf{v} = \langle v_1, v_2, v_3 \rangle$ , the cross product is the vector $\langle u_2v_3 - u_3v_2,\; u_3v_1 - u_1v_3,\; u_1v_2 - u_2v_1 \rangle$ . It is defined for three-dimensional vectors, and the result is again a vector in three dimensions.

Its direction is perpendicular to both $\mathbf{u}$ and $\mathbf{v}$ , chosen by the right-hand rule: curl the fingers of your right hand from $\mathbf{u}$ towards $\mathbf{v}$ and your thumb points along $\mathbf{u} \times \mathbf{v}$ . Its length is $|\mathbf{u}|\,|\mathbf{v}|\sin\theta$ , the area of the parallelogram the two vectors span.

How to use the Cross Product Calculator

Enter the three components of the first vector separated by commas, for example 1, 2, 3.
Enter the three components of the second vector.
Press Calculate.
Read the i, j and k components and the resulting vector.
Use the magnitude step for areas and the orthogonality check to confirm the result.

Formulas

Determinant form

\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}

Magnitude

|\mathbf{u} \times \mathbf{v}| = |\mathbf{u}|\,|\mathbf{v}|\sin\theta

Anticommutativity

\mathbf{v} \times \mathbf{u} = -(\mathbf{u} \times \mathbf{v})

Triangle area

A = \tfrac12\,|\mathbf{u} \times \mathbf{v}|

Worked example: ⟨1,2,3⟩ × ⟨4,5,6⟩

The example crosses $\mathbf{u} = \langle 1, 2, 3 \rangle$ with $\mathbf{v} = \langle 4, 5, 6 \rangle$ . The $\mathbf{i}$ component is $2 \cdot 6 - 3 \cdot 5 = -3$ ; the $\mathbf{j}$ component is $-(1 \cdot 6 - 3 \cdot 4) = 6$ ; the $\mathbf{k}$ component is $1 \cdot 5 - 2 \cdot 4 = -3$ .

So $\mathbf{u} \times \mathbf{v} = \langle -3, 6, -3 \rangle$ , with magnitude $\sqrt{54} = 3\sqrt6 \approx 7.348$ , the area of the parallelogram spanned by the two vectors. Its dot product with each input is 0, confirming that it is perpendicular to both.

The solution below is generated by the same engine that powers the calculator above.

Final answer

u \times v = \langle -3, 6, -3 \rangle

How we read your problem

Cross product of two 3D vectors.

\langle 1, 2, 3 \rangle \times \langle 4, 5, 6 \rangle

Step-by-step solution

Write the determinant with unit vectors i, j, k in the first row
$u \times v = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{vmatrix}$
i component
$(2)(6) - (3)(5) = -3$
j component (note the minus sign in the expansion)
$-\big((1)(6) - (3)(4)\big) = 6$
k component
$(1)(5) - (2)(4) = -3$
Magnitude = area of the parallelogram spanned by u and v
$|u \times v| = \sqrt{54} = 3\sqrt{6} \approx 7.348469228$
Check orthogonality
$(u\times v) u = 0,\quad (u\times v) v = 0$

Formulas used

Cross product
$u \times v = \langle u_2v_3 - u_3v_2,\ u_3v_1 - u_1v_3,\ u_1v_2 - u_2v_1 \rangle$
Magnitude
$|u \times v| = |u||v|\sin\theta$

The result is perpendicular to both u and v; its length 3√6 ≈ 7.348469228 is the area of the parallelogram they span (the triangle has half that area).

Verification

The result is perpendicular to both input vectors (both dot products are 0).

Detailed explanation

The determinant method

Writing the unit vectors in the first row of a $3 \times 3$ determinant and expanding along that row is the easiest way to remember the formula. Each component is a $2 \times 2$ determinant formed from the columns not belonging to that unit vector. The middle term carries a minus sign from the cofactor pattern $+, -, +$ , which is why the calculator labels it explicitly.

The cross product distributes over addition, $\mathbf{u} \times (\mathbf{v} + \mathbf{w}) = \mathbf{u} \times \mathbf{v} + \mathbf{u} \times \mathbf{w}$ , and scaling either input scales the result. It is not associative, however: $(\mathbf{u} \times \mathbf{v}) \times \mathbf{w}$ and $\mathbf{u} \times (\mathbf{v} \times \mathbf{w})$ are generally different vectors. For flat vectors such as $\langle 1, 2, 0 \rangle$ and $\langle 3, 4, 0 \rangle$ the result $\langle 0, 0, -2 \rangle$ points straight down the $z$ -axis, and its length 2 is the area of the parallelogram in the plane.

Geometric meaning

The magnitude of the cross product equals the area of the parallelogram with sides $\mathbf{u}$ and $\mathbf{v}$ , so half of it is the area of the triangle they form. For a triangle with vertices $P$ , $Q$ and $R$ , cross $\overrightarrow{PQ}$ with $\overrightarrow{PR}$ and halve the length.

If the vectors are parallel, the parallelogram collapses and the cross product is the zero vector. For $\langle 1, 2, 3 \rangle$ and $\langle 2, 4, 6 \rangle$ the calculator returns $\langle 0, 0, 0 \rangle$ , which is a quick test for parallel vectors.

Normals, planes and physics

Crossing two direction vectors that lie in a plane gives a normal vector to the plane, and with one point on the plane you can write its equation $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$ . In computer graphics the same calculation gives surface normals for lighting.

In physics, torque is $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$ and the magnetic force on a moving charge is $q\,\mathbf{v} \times \mathbf{B}$ . In each case the result points perpendicular to the plane of the two inputs, as the right-hand rule predicts.

Common mistakes

Forgetting the minus sign on the $\mathbf{j}$ component.
Assuming $\mathbf{u} \times \mathbf{v} = \mathbf{v} \times \mathbf{u}$ ; reversing the order reverses the direction.
Applying the cross product to two-dimensional vectors without first adding a zero third component.
Confusing the cross product with the dot product, which returns a number.
Forgetting to halve the magnitude when finding the area of a triangle.

Tips

Check your answer by dotting it with each input; both results must be 0.
Use $\mathbf{i} \times \mathbf{j} = \mathbf{k}$ , $\mathbf{j} \times \mathbf{k} = \mathbf{i}$ and $\mathbf{k} \times \mathbf{i} = \mathbf{j}$ as a memory aid.
For 2D vectors, enter 0 as the third component; the answer then points along the $z$ -axis.
Use the dot product calculator to find the angle between the vectors.
Use the determinant calculator to compute the scalar triple product $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$ as a $3 \times 3$ determinant.
Sketch the two vectors and apply the right-hand rule to predict the sign of each component before you calculate.

Frequently asked questions

How do you calculate the cross product?

Place i, j and k in the first row of a 3×3 determinant and the two vectors in the next rows, then expand along the first row. Remember the minus sign on the j component.

What does the magnitude of the cross product represent?

The area of the parallelogram spanned by the two vectors. Half of it is the area of the triangle they form.

Is the cross product commutative?

No. Swapping the order reverses the direction: v × u = −(u × v).

Can I take the cross product of 2D vectors?

Not directly; the cross product is defined for three components. Add 0 as the third component of each vector and the result will point along the z-axis.

What does a zero cross product mean?

The vectors are parallel (or one is the zero vector), so they span no area.

How do I check a cross product?

Take its dot product with each original vector. Both results should be zero because the cross product is perpendicular to both inputs. The calculator does this check for you.

How do I find a normal vector to a plane?

Take two non-parallel vectors that lie in the plane and compute their cross product. The result is perpendicular to the plane.

Compute your cross product

Enter two 3D vectors above to get their cross product, its magnitude and the orthogonality check.

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